To clarify how Python handles two equal objects

[Chris Angelico]

On Wed, 11 Jan 2023 at 07:14, Jen Kris via Python-list
<python-list at python.org> wrote:
>
> I am writing a spot speedup in assembly language for a short but computation-intensive Python loop, and I discovered something about Python array handling that I would like to clarify.
>
> For a simplified example, I created a matrix mx1 and assigned the array arr1 to the third row of the matrix:
>
> mx1 = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
> arr1 = mx1[2]
>
> The pointers to these are now the same:
>
> ida = id(mx1[2]) - 140260325306880
> idb = id(arr1) - 140260325306880
>
> That’s great because when I encounter this in assembly or C, I can just borrow the pointer to row 3 for the array arr1, on the assumption that they will continue to point to the same object.  Then when I do any math operations in arr1 it will be reflected in both arrays because they are now pointing to the same array:
>

That's not an optimization; what you've done is set arr1 to be a
reference to that object.

> But on the next iteration we assign arr1 to something else:
>
> arr1 = [ 10, 11, 12 ]
> idc = id(arr1) – 140260325308160
> idd = id(mx1[2]) – 140260325306880
>
> Now arr1 is no longer equal to mx1[2], and any subsequent operations in arr1 will not affect mx1.

Yep, you have just set arr1 to be a completely different object.

> So where I’m rewriting some Python code in a low level language, I can’t assume that the two objects are equal because that equality will not remain if either is reassigned.  So if I do some operation on one array I have to conform the two arrays for as long as they remain equal, I can’t just do it in one operation because I can’t rely on the objects remaining equal.
>
> Is my understanding of this correct?  Is there anything I’m missing?
>

Assignment in Python is a matter of object references. It's not
"conform them as long as they remain equal". You'll have to think in
terms of object references the entire way.

ChrisA